On Feb 23, 12:39 pm, Martin Matysiak <[email protected]> wrote:
> I think it depends on the image you are using. Is the scale linear in both
> directions (vertical and horizontal)? Then it should be simple mathematics.
>
> dX = X2 - X1 = e.g. 42
> dLon = Lon2 - Lon1 = e.g. 21
>
> LonPerPixel = dLon / dX = 0.5
>
> That means, each pixel (in a linear scaled image!) represents a longitude
> step of LonPerPixel degrees. You do the same for LatPerPixel (with Y and Lat,
> of course) and then it should be pretty simple to get the X/Y coordinates of
> the third point. Just get the Lat/Lon differences between the third and the
> first or second point, divide the differences through LonPerPixel and
> LatPerPixel and you know how much pixels you have to add to your first resp.
> second point in order to get the coordinates of the third one!
>
> However, this could fail if P1 and P2 are either horizontally or vertically
> on the same line (as one of the differences would equal zero) or the image
> is scaled in any other way.
>
> Regards,
>
>   Martin Matysiak
>   homepage <https://martin-matysiak.de> | blog<http://blog.martin-matysiak.de>

Google's Mercator projection is not linear in the vertical direction.
It is misleading to suggest it is.

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